Proof of x and y int of a line

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The discussion focuses on proving that the sum of the x and y intercepts of tangents to the line defined by the equation √x + √y = c equals a constant c. Participants express confusion about the correct approach to find the intercepts and clarify the use of the point/slope form of the tangent line equation. There is a significant error identified regarding the derivative, where the correct relationship between x and y is emphasized. The need for clear definitions of variables x1, y1, x2, and y2 is also highlighted. Overall, the conversation centers on resolving misunderstandings in calculus concepts related to the problem.
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Homework Statement


prove that \sqrt{x} + \sqrt{y} = c
Show that the sum of the x and y intercepts of any tangents to the line above = c (some positive constant).


Homework Equations


y1 - y2 = m(x1 - x2)
dy/dx(for this problem) = -\sqrt{y}/\sqrt{x}


The Attempt at a Solution


So I get the slope, as written above, and put it into a point/slope equation, but where from here? When I try to solve for y = 0 and x = 0 I always have y2 and x2 left, I think I might just be doing something completely wrong, I haven't done something like this for a while. Is this even the right direction? Solve for the y and x int by making the opposite 0 in the equation, and then try to get the results, (the two interecepts) to add up to be \sqrt{x} + \sqrt{y}
 
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First problem, I don't know what "x1", "y1", "x2", and "y2" are since you don't say. You mean, I think, that the equation of the tangent line at (x_1, y_1) is y- y_1= m(x- x_1). <br /> <br /> But your real problem is that the derivative of y with respect to x is NOT -\sqrt{y}/\sqrt{x}. You have x and y reversed.
 
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